Find pH and concentration of HSO3- and SO3-2 in .05 M NaHSO3.?

Kw = 1.0 × 10⁻¹⁴

Refer to: https://www.thoughtco.com/acids-and-bases-weak-acid-ka-values-603973
Kₐ₁ for H₂SO₃ = 1.4 × 10⁻²
Kₐ₂ for H₂SO₃ = 6.3 × 10⁻⁸
(Ka values from different sources may be slightly different.)

HSO₃⁻ ion is amphoteric, which can act as an acid (1) and a base (2):
(1): HSO₃⁻(aq) + H₂O(ℓ) ⇌ SO₃²⁻(aq) + H₃O⁺(aq) … Ka = Kₐ₂ = 6.3 × 10⁻⁸
(2): HSO₃⁻(aq) + H₂O(ℓ) ⇌ H₂SO₄(aq) + OH⁻(aq) … Kb = Kw/Kₐ₁ = 7.1 × 10⁻¹³
As Ka ≫ Kb, we consider reaction (1) only.

________ HSO₃⁻(aq) + H₂O(ℓ) ⇌ SO₃²⁻(aq) + H₃O⁺(aq) ___ Ka = 6.3 × 10⁻⁸
Initial: ____ 0.05 M ___________ 0 M _____ 0 M
Change: ____ -y M ___________ +y M ____ +y M
Eqm: __ (0.05 - y) M ___________ y M ____ y M

As Ka is very small, the dissociation of HSO₃⁻ is to a very small extent.
It can be assumed that 0.05 ≫ y
Hence, equilibrium [HSO₃⁻] = (0.05 - y) M ≈ 0.05 M

At equilibrium:
Ka = [SO₃²⁻] [H₃O⁺] / [HSO₃⁻]
6.3 × 10⁻⁸ = y² / 0.05
y = √(0.05 × 6.3 × 10⁻⁸) = 5.61 × 10⁻⁵

pH = -log[H₃O⁺] = -log(5.61 × 10⁻⁵) = 4.25[HSO₃⁻] = 0.05 M
[SO₃²⁻] = 5.61 × 10⁻⁵ M